A Spring is stretched by 0.20 m when mass of 0.50 kg is suspended. When a mass of 0.25 kg is suspended, then its period of oscillation will be \[\left( {g\, = 10{\kern 1pt} m/{s^2}} \right)\].
A. 0.328 sec
B. 0.628 sec
C. 0.137 sec
D. 1.00 sec

Hint: Force on a spring of constant k is equal to the product of spring constant and length by which spring is extended. Therefore, the spring constant is directly proportional to the Force applied on a spring.

Formula used:
Time period \[T = 2\Pi \sqrt {\dfrac{m}{k}} \]
Where m = mass suspended to spring and k = spring constant.
Force F =- kx = mg. Here x is length of extension and g is acceleration due to gravity.

Complete step by step solution:
Given here is a spring which is stretched by 0.20 m when 0.50 kg mass is suspended by it we have to find its period of oscillation when the suspended mass is 0.25 kg.
As we know for a spring mass system of constant is k and suspended mass m, its time period of oscillation is given by,
\[T = 2\Pi \sqrt {\dfrac{m}{k}} \,........(1)\]
To find the time period of oscillation we need the value of spring constant k.
Force on the spring when 0.50 kg mass leads to a stretch of 0.20 m is given by kx.
As force is also given by equation F = mg and on equating kx with mg we get,
\[k = \dfrac{{mg}}{x}\,.......(2)\]
Substituting m = 0.50 kg, \[g\, = 10{\kern 1pt} m/{s^2}\]and x = 0.20 m in equation (2) we get,
\[k = \dfrac{{0.50\,kg \times 10\,m/{s^2}}}{{0.20\,m}} \Rightarrow k = \,25\,N/m\]
Now, as the value of spring constant is known, the time period of oscillation for mass 0.25 kg suspended by the spring can be calculated.
Substituting, \[k = \,25\,N/m\]and m = 0.25 kg in equation (1) we get,
\[T = 2\Pi \sqrt {\dfrac{{0.25}}{{25}}} \, \Rightarrow T = 2\Pi \left( {\dfrac{{0.5}}{5}} \right)\]
On solving the above equation value of T is given by,
T = 0.628 sec
Hence, Time period of oscillation when 0.25 kg mass is suspended by spring is 0.628 sec.
Therefore, option B is the correct option.

Note: Negative sign in F = -kx signifies the restoring force of the spring as the force exerted by the spring is in direction opposite to the direction of displacement, but in case of stretching of spring, force on the spring will be positive as the force and displacement are in same direction.